# Proving $\int_{0}^{1}xf(x)dx \leq \frac{2}{3}\int_{0}^{1}f(x)dx$ for all concave functions $f: [0,1]\rightarrow [0,\infty)$

I'm trying to prove the inequality

$$\int_{0}^{1}xf(x)dx \leq \frac{2}{3}\int_{0}^{1}f(x)dx$$

for all continuous concave functions $$f: [0,1]\rightarrow [0,\infty)$$. I've been working on this for a while and would just love a hint if anyone can see something obvious that I'm missing.

What I've tried is observing that $$\frac{2}{3} = \int_{0}^{1}\sqrt{x}dx$$ and trying to use Holder's inequality with $$p=q=2$$ on the integral $$\int_{0}^{1}xf(x)$$ but the squaring and the integral seem to be in the wrong order. I think there might be something more to this idea since that's the only way I know off the top of my head that can split an integral of a product into a product of the integral at the cost of inequality. However, I can't see how concavity comes into play.

Something else I've atempted is observing that because of the concavity of $$f(x)$$ we have for all $$x \in [0,1]$$ that $$xf(x) + (1-x)f(0) \leq f(x^2 + 0 (1-x)) = f(x^2)$$

so that $$\int_{0}^{1}xf(x) + (1-x)f(0) \leq \int_{0}^{1}f(x^2) \leq f\left(\int_{0}^{1}x^2dx\right) = f(1/3)$$ where the second to last inequality is Jenson's inequality (since $$f(x)$$ is concave). Evaluating the first integral as much as possible gave $$\int_{0}^{1}xf(x)dx + f(0)/2 \leq f(1/3)$$ but the above result feels incorrect somehow eventhough I feel fairly confident in every step.

We just want to show $$\int_0^1 (x-2/3)f(x)dx \leq 0$$ We have \begin{align} \int_0^1 (x-2/3)f(x)dx &= \int_0^{2/3} \underbrace{(x-2/3)}_{\leq 0}f(x)dx + \int_{2/3}^1 \underbrace{(x-2/3)}_{\geq 0}f(x)dx \\ &\leq \int_0^{2/3}(x-2/3)L(x)dx + \int_{2/3}^1(x-2/3)U(x)dx \end{align} where $$L(x)$$ and $$U(x)$$ are any lower-bound and upper-bound functions over their respective intervals, so that \begin{align} L(x) &\leq f(x) \quad \forall x \in [0,2/3] \quad \mbox{(Eq. 1)}\\ U(x) &\geq f(x) \quad \forall x \in [2/3,1] \quad \mbox{(Eq. 2)} \end{align} But by concavity and nonnegativity of $$f$$, the linear function $$L(x) = U(x) = x\frac{f(2/3)}{2/3} \quad \forall x \in \mathbb{R}$$ satisfies (Eq. 1) and (Eq. 2). (Just draw a picture and note that a concave function lies above its chords.) So $$\int_0^1 (x-2/3)f(x)dx \leq \int_0^1(x-2/3)x\frac{f(2/3)}{2/3}dx = 0$$ The bound is tight because it holds for the linear function $$f(x)=x$$.